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1 year ago

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Samantha is making a vegetable soup that contains carrots, beans, water, salt, pepper, bits of ham, and onions. the soup has to

cook for 4 hours. while the soup is cooking, what will happen to some of the minerals that are in its ingredients?

1 answer:

murzikaleks [220]1 year ago

8 0

<span>While the soup is cooking, some of the minerals that are in its ingredients will flow out and mixed or react with the compounds in the water. It follows the law of osmosis where the flow is from a higher concentration to a lower concentration. That is why, at the end of cooking time, the water which was originally bland in taste, will taste like meat and veggie broth.</span>

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According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams

gogolik [260]

Answer:

$m_{HClO_3}=12.7gHClO_3$

Explanation:

Hello,

Considering the reaction:

$3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3$

The molar masses of chlorine and chloric acid are:

$M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol$

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

$m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3$

Best regards.

4 0

2 years ago

Radioactive elements decay at a know rate known as half-life. Choose all of the correct statements concerning half-life. Carbon-

9966 [12]

Explanation:

Half life is simply the amount of time it takes for half of a substance to decompose.

Options;

- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.

- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000

- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)

- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)

- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)

6 0

1 year ago

Read 2 more answers

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

A 50.6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45.0 grams of hydrochloric acid (HCl). What is the theoreti

zysi [14]

<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)

grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required

Since there are only 45.0 grams HCl, then HCl is the limiting reactant.

theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>

7 0

2 years ago

Read 2 more answers

What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution

EastWind [94]

Answer:

The volume required for complete neutralize is 32.29 mL

Explanation:

The computation of the volume required for complete neutralize is shown below:

As we know that, the balanced equation is

$Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O$

Now

The number of moles of $Ba(OH)_2$ = n_1 = 1

And, the number of moles of Hcl = n_2 = 2

Therefore

The equation i.e. to be used to find out the volume is given below:

$\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$

$V_2 = \frac{M_1V_1}{n_1} \times \frac{n_2}{M_2} \\\\ = \frac{0.2350 \times 23.55}{1} \times \frac{2}{0.3428} \\\\ = \frac{11.0685}{0.3428}$

= 32.29 mL

Hence, the volume is 32.29mL

6 0

2 years ago